Question 946175
An object is projected vertically from the top of a building with an initial velocity of 128ft/sec. It's [sic] distance is s(t) in feet above the ground after t seconds is given by the equation.

S(t)= -16t^2 + 128t +80

a. Find its maximum distance above the ground.
Max height is the vertex of the parabola at t - -b/2a
t = -128/-32 = 4 seconds
s(4) = -16*16 + 128*4 + 80 = 336 feet
=======================
b. Find the height of the building.
It's s(0) = 80 feet